Answer
$2.87\times 10^{-17}\;J$
Work Step by Step
The normalized wavefunction $\psi_n(x)$ for an electron in an infinite, one-dimensional potential well with length $L$ along an $x$ axis are given by
$\psi_n(x)=\sqrt {\frac{2}{L}}\sin\Big(\frac{n\pi}{L}x\Big),\;\;\;\text{for}\;n=1,2,3....,$
where $n$ is the quantum number.
Now, the probability that the electron will be detected in the well is given by
$\rho(x)=\psi^2_n(x)$
or, $\rho(x)={\frac{2}{L}}\sin^2\Big(\frac{n\pi}{L}x\Big)$
For the given potential well, the electron’s probability density is zero at $x=0.300L=\frac{3L}{10}$, and $x=0.400L=\frac{4L}{10}$
Thus,
$\rho(\frac{3L}{10})={\frac{2}{L}}\sin^2\Big(\frac{3n\pi}{10}\Big)=0\implies\frac{3n\pi}{10}=m\pi\implies m=\frac{3n}{10}$
and
$\rho(\frac{4L}{10})={\frac{2}{L}}\sin^2\Big(\frac{4n\pi}{10}\Big)=0\implies\frac{4n\pi}{10}=m\pi\implies m=\frac{4n}{10}$
In above relations, $m$ is integer and $n$ is positive integer. Therefore, the solution will be $n=10m$.
The electron’s probability density is non-zero in-between $x=0.300L$, and $x=0.400L$. Therefore $m=1$ and $n=10$
An electron confined to an infinite potential well can exist in only certain discrete states. If the well is one-dimensional with length $L$, the energies associated with these quantum states are
$E_n=\Big(\frac{h^2}{8mL^2}\Big)n^2,\;\;\;\text{for}\;n=1,2,3....,$
where $m$ is the electron mass and $n$ is a quantum number.
The lowest energy corresponds to the ground state $(n=1)$ is,
$E_1=\frac{h^2}{8mL^2}$
Thus the energies associated with the quantum states of an electron confined to an infinite potential well can be written as
$E_n=n^2E_1$
If the electron then jumps to the next lower energy level $(n=9)$ by emitting light, the change in the electron’s energy is given by
$\Delta E=10^2E_1-9^2E_1=19E_1$
Substituting the given values,
$E_1=\frac{(6.63\times10^{-34})^2}{8\times 9.1\times10^{-31}(200\times 10^{-12})^2}$
or, $E_1\approx 1.51\times 10^{-18}\;J$
Thus, $\Delta E=19\times1.51\times 10^{-18}\;J\approx 2.87\times 10^{-17}\;J$
